A sequence of reactions is shown below starting from an optically active compound P. $$\Large{\ce{\underset{\ce{C4H8O3}}{P}->[SOCl2]\underset{\ce{C4H6Cl2O}}{Q}->[C2H5NH2]\underset{\ce{C6H12ClNO}}{R}}}$$ P does not react with 2,4-dinitrophenylhydrazine (2,4-DNP) to form a hydrazone. Q reacts very rapidly with one molar equivalent of ethylamine at low temperature to form R. Select the most appropriate statement(s). (A) Q as well as R are optically active. (B) Q on heating with an excess of ethylamine in the presence of a base gives a basic compound. (C) R is a basic compound and can react with an acid to form a salt. (D) An optically inactive isomer of P on heating can form compound S $(\ce{C4H6O2})$ which does not react with thionyl chloride.

My approach:

P is hydroxy butanoic acid. In Q the hydroxyl groups gets replaced by chlorine. I have confusion about the structure of R.

I feel Option (A) to be correct. I am doubtful about (D). And have no idea about (B) and (C).

  • 1
    $\begingroup$ Some hint: 1. Why you think there is optical activity involved? It is not mentioned in the question; 2. Check the thionyl chloride reaction to find out how hydroxy carboxylic acid will react with it. It is more than you have thought; 3. Amine is basic but amide is not; 4. Alkyl chloride will not react with amine easily as room temperature, you need more reactive specie to react with amine "very rapidly". $\endgroup$
    – Coconut
    May 12, 2014 at 16:01

1 Answer 1


I think your guess for P is correct, we just don't know whether it is 2- or 3-hydroxy-butanoic acid. Reaction of thionyl chloride with the hydroxyl group in the alkyl chain can proceed with retention or inversion of configuration depending on reaction conditions, but the point is, if P is optically active, then Q will be optically active too (so A seems correct). Reaction of Q, an acid chloride, with ethyl amine will produce the corresponding ethyl amide, R. Since amides are basic (they can form salts with acids), I think B and C are true too. An optically inactive isomer of P would be 4-hydroxy-butanoic acid. On heating it would dehydrate and form a stable lactone (S) that would not react with thionyl chloride, so I think D is true also. I just wonder if we're missing something that would allow us to determine if P is the 2- or 3- isomer?

Update: To be complete and accurate let's update. Based on the comments below (particularly JA's) I was right about "B" (but for the wrong reason, I thought the amide would be basic enough, but it is not; B is true because, as JA points out, heating with excess amine will convert P into an amino amide (I've labeled it as X) which is basic). I was wrong about C, R is not basic enough to form a salt. I still don't know if P is 2- or 3-hydroxy-butanoic acid, but I've added a picture based on the 3-hydroxy isomer to try and summarize the chemistry.

enter image description here

  • $\begingroup$ A lot of mistake here. 1. Existence of chiral center does not mean it is optically active, since it can be racemic; 2. Amides are basic? Are you kidding? 3. The purpose of this question is to decide whether the statements is true or false, not to find the structure. And you do not have to. $\endgroup$
    – Coconut
    May 12, 2014 at 16:08
  • $\begingroup$ Amides are very weak bases compared to amines, because of the electron-withdrawing carbonyl group and the nitrogen lone pair being involved in resonance, so I have some doubt that (C) is true. However, (B) is true because the chloride in the alkyl chain can be substituted by ethylamine under these conditions. $\endgroup$ May 12, 2014 at 16:18
  • $\begingroup$ @Ian Fang 1. In the first sentence of the problem it says, "optically active compound P", 2) amides are weakly basic, 3) finding the structure helped "me" figure things out - didn't say everyone had to do it that way. $\endgroup$
    – ron
    May 12, 2014 at 16:25
  • $\begingroup$ @Jannis Andreska. Your suspicion about (C) is correct. (C) is false. But what is S? I know the reaction between acid chloride and ethanamine in cold. But what happens in the reaction specified in (D)? $\endgroup$
    – Rudstar
    May 12, 2014 at 18:32
  • $\begingroup$ @Rudstar Like Ron already suggested, the optically inactive isomer of P is 4-hydroxybutanoic acid, because it does not contain a stereocenter. On heating, this compound can form $\gamma$-butyrolactone by intramolecular esterification, which would be compound S. This lactone does not react with thionyl chloride. $\endgroup$ May 12, 2014 at 18:44

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